OR since a^2-b^2=(a+b)(a-b) then that could be written as
(2001+2000)(2001-2000)+(1999+1998)(1999-1998)+...+(3+2)(3-2)+1
=(2001+2000)(1)+(1999+1998)(1)+...+(3+2)(1)+1
=2001+2000+1999+1998+...+3+2+1
=1+2+3+.....+1998+1999+2000+2001
this is an arithmetic series now we can use the formula
\[s=(n/2)(a_{1}+a_{n})\]
where a sub n is the last term of the series and and a sub 1 is the first term of the series and n is the number of terms in the series so let's compute for the sum...
s=(2001/2)(1+2001)
s=(2001/2)(2002)
s=(2001)(1001)
s=2003001 is the sum :D